1/ \(3x^2-2x=0\Leftrightarrow x\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy .....................
2/ \(4x^2-4x+\frac{1}{2}=0\)
\(\Delta=b'^2-ac=2^2-\frac{4.1}{2}=2\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\frac{2+\sqrt{2}}{4}\\x_2=\frac{2-\sqrt{2}}{4}\end{matrix}\right.\)
Vậy ........................
3/ \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy ......................
a) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-2}{3}\right\}\)
b) Ta có: \(4x^2-4x+\frac{1}{2}=0\)
\(\Leftrightarrow4x^2-4x+1-\frac{1}{2}=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\frac{1}{2}=0\)
\(\Leftrightarrow\left(2x-1\right)^2=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{\frac{1}{2}}\\2x-1=-\sqrt{\frac{1}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{\frac{1}{2}}+1\\2x=-\sqrt{\frac{1}{2}}+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{2}}{4}\\x=\frac{2-\sqrt{2}}{4}\end{matrix}\right.\)
Vậy: \(x=\frac{2\pm\sqrt{2}}{4}\)
1, 3x2+2x=0
=>x(3x+2)=0
=>x=0 hoặc x=\(\frac{-2}{3}\)
2, nhấn shift 5, 3 ra ngay kết quả
3, (x-1)(x2+5x-2)-(x3-1)=0
=>(x-1)(x2+5x-2)-(x-1)(x2+x+1)=0
=>(x-1)(x2+5x-2-x2-x-1)=0
=>x=1 hoặc x=\(\frac{1}{4}\)
a, 3x2 + 2x = 0
\(\Leftrightarrow\) x(3x + 2) = 0
\(\Leftrightarrow\) x = 0 hoặc 3x + 2 = 0
\(\Leftrightarrow\) x = 0 và x = \(\frac{-2}{3}\)
Vậy S = {0; \(\frac{-2}{3}\)}
b, 4x2 - 4x + \(\frac{1}{2}\) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 - \(\frac{1}{2}\) = 0
\(\Leftrightarrow\) (2x - 1)2 - \(\frac{1}{2}\) = 0
\(\Leftrightarrow\) (2x - 1 - \(\sqrt{\frac{1}{2}}\))(2x - 1 + \(\sqrt{\frac{1}{2}}\)) = 0
\(\Leftrightarrow\) 2x - 1 - \(\sqrt{\frac{1}{2}}\) = 0 hoặc 2x - 1 + \(\sqrt{\frac{1}{2}}\) = 0
\(\Leftrightarrow\) x = \(\frac{\sqrt{\frac{1}{2}}+1}{2}\) và x = \(\frac{-\sqrt{\frac{1}{2}}+1}{2}\)
c, (x - 1)(x2 + 5x - 2) - (x3 - 1) = 0
\(\Leftrightarrow\) (x - 1)(x2 + 5x - 2) - (x - 1)(x2 + x + 1) = 0
\(\Leftrightarrow\) (x - 1)(x2 + 5x - 2 - x2 - x - 1) = 0
\(\Leftrightarrow\) (x - 1)(4x - 3) = 0
\(\Leftrightarrow\) x - 1 = 0 hoặc 4x - 3 = 0
\(\Leftrightarrow\) x = 1 và x = \(\frac{3}{4}\)
Vậy S = {1; \(\frac{3}{4}\)}
Hình như phần b lỗi rùi, phải là \(\frac{1}{4}\) mới đúng chớ, chứ để \(\frac{1}{2}\) ko ra được đâu
Chúc bn học tốt!!
